广西钦州市2019-2020学年高二下学期期末教学质量监测数学(文)试题扫描版含答案

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钦州市2020年春季学期教学质量监测参考将答案高二数学(文科)一、选择题答案:(每小题5分,共60分)题号123456789101112答案BBDDCCDABBCD二、填空题答案:(每小题5分,共20分)13

.(1,2);14.5;15.2;16.8月份利润最低(或“3月份和10月份利润最高”等其他正确结论)三、解答题:本大题共6题,共70分.解答应写出文字说明、证明过程或演算步骤.17.证明:要证511313只须

证22511313········································································2分只须证525511323913···················

···················································.4分只须证5539···························································

····································6分只须证5539·································································································8分

因为5539成立所以511313·································································10分18.解:(1)所有感染病毒的小白鼠共有50只,其中注射疫苗的共有y只,∴1505yP···

··········································································································2分∴10,501040,yx402060,301040,mn

·····························4分∴22列联表如下:未感染病毒感染病毒总计未注射疫苗204060注射疫苗301040总计5050100·································

····································································6分(2)∵22100(20103040)1005016.6675050604063K·····

·······························10分∵16.6676.635∴有99%把握认为注射此种疫苗对预防新型冠状病毒有效.························12分19.解:(1)(

1)∵cosx,siny,·························································2分∴1C的极坐标方程为sin2.·································

······························4分(2)∵直线3C的极坐标方程为6πR∴124sin6π,························································

············································7分24sin()23,66·················································

·········································10分∴21423MN···········································································

········12分20.解:∵()|2|1fxxx,()7fx∴①当1x时,原不等式可化为217,x解得[3,1]x;··························3分②当12x时,原不等式可化为37,解得(1,2)x

;···································6分③当2x时,不等式可化为217,x解得[2,4]x;··········································9分综上,原不等式的解集为

[3,4]···········································································12分21.解:(1)∵32cos,2sinxy∴2222(3)2cos2sin4xy

∴曲线C的直角坐标方程22(3)4xy·······························································4分(2)∵直线l的参数方程为222,2

2xtyt(t为参数)∴直线l与x轴交于点(2,0)P····································································

············6分将222,22xtyt代入圆C的方程22(3)4xy,整理得2230,tt122,tt12·3tt························

·············································································8分由参数的几何意义得1212PMPNtttt21212()4tt

tt2(2)4(3)14.∴14MNPMPN·········································································12分22.解:(1)由表中数据得

,23453.54x,26394954424y,∴41421ˆ9.4iiiiixxyybxx,429.43.59.1aybx,∴回归方程为9.49.1yx.············

·······································································6分(2)年资金投入量为7.5万元时,9.47.59.179.6y(万元);··················

··8分(3)解9.49.1100x,······························································································10分得90.

99.4x∵90.99.679.4∴若年市场销售额超过100万,那么年资金投入量至少要9.67万元.··········12分

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